$2\sqrt{x} + \sqrt{3}$
How do I simplify this?
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9It can’t be simplified any further. – Brian M. Scott Oct 03 '11 at 23:25
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Check the edit. I left something out. – David G Oct 03 '11 at 23:28
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2That doesn't make it any better. – joriki Oct 03 '11 at 23:32
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2Brian's comment still applies, even after your edit. – J. M. ain't a mathematician Oct 03 '11 at 23:32
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Okay well thanks anyway. – David G Oct 03 '11 at 23:37
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Are you perhaps intended to solve the equation $2\sqrt{x} + \sqrt{3} = 0$? – Austin Mohr Oct 04 '11 at 00:20
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4What's not simple about that? – davidlowryduda Oct 04 '11 at 00:47
2 Answers
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This seems a simpler derivation to me:
Squaring $a \sqrt x + \sqrt b$ we get $a^2 x + b + 2a \sqrt{bx}$ so $$ a \sqrt x + \sqrt b = \sqrt{a^2 x + b + 2a \sqrt{bx}}. $$
This seems to be a more complicated result to me.
I know, it's all about me.
marty cohen
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Maybe you are searching for something similar to this:
$$\sqrt{a{^+ _-}\sqrt{b}}=\sqrt{\frac{a+c}{2}}{^+ _-}\sqrt{\frac{a-c}{2}}$$
$$c=\sqrt{a^2-b}$$
A example:
$\sqrt{5{^+_-}\sqrt{24}}=\sqrt{\frac{5+1}{2}}{^+_-}\sqrt{\frac{5-1}{2}}=\sqrt{3}{^+_-}\sqrt{2}$
In your problem:
$2\sqrt{x}+\sqrt{3}=\sqrt{3}+\sqrt{4x}$
$a+c=2\times3$
$a-c=2\times4x$
So,
$a=4x+3$
$c=3-4x$
$b=a^2-c^2=(a-c)(a+c)=8x\times6=48x=3x\times4^2$
The final is:
$$2\sqrt{x}+\sqrt{3}=\sqrt{4x+3+4\sqrt{3x}}$$
But it needs a better investigation of the existence conditions in these calculus.
GarouDan
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5I haven't followed the details of the calculations, but how is the final result supposed to be simpler than $2\sqrt{x} + \sqrt{3}$? – t.b. Nov 11 '11 at 01:36
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1It's no simpler (I wrote Maybe you are searching for something similar to this:), it's just a transformation and maybe was he is looking for. If he tells not I'll delete if none think this useful. – GarouDan Nov 16 '11 at 13:26
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